I've also found what it converges to, the golden ratio w/ a calculator. How would I go about this not using a calculator? Usually one would just consider the infinite case but here I'm not sure how to do that.

What I'm saying is since each of the subsequent terms depends on the one before it... that sort of seems like magic to me. when you go to infinity there negligible difference between the n+1th and nth term in the sequence, but still...

Even with a(n)=[tex]\sqrt{1+a(n)}[/tex], what would I take a(n) to obtain the Golden ratio?

Sorry if I'm not making sense; I'm confused as to how making this equivalence helps us find the limit; we still don't have an (n) to take the limit with respect to...

I have to disagree with tiny-tim here, at least with the way he said it. Saying "put n= n+1" is exactly the same as saying 0= 1! What he meant, I'm sure, is that the sequence {a(n)} necessarily has the same limit as the sequence {a(n+1)}, the same sequence, "shifted" by 1. That leads to what evail is saying:
[tex]\lim_{n\rightarrow\infty} a(n)= \sqrt{1+ \lim_{x\rightarrow\infty} a(n+1)}[/tex]
so if the limit is "[itex]l[/itex]", [itex]l= \sqrt{1+ l}[/itex] which has the solutions evail shows.

Of course, the limit is one of [itex](1+ \sqrt{5})/2[/itex] or [itex](1- \sqrt{5})/2[/itex]. Do you see how to determine which one? a(1)= 1>0 and it is easy to see that this sequence is increasing. The limit can't be negative!

I have to disagree with tiny-tim here, at least with the way he said it. Saying "put n= n+1" is exactly the same as saying 0= 1! What he meant, I'm sure, is that the sequence {a(n)} necessarily has the same limit as the sequence {a(n+1)}, the same sequence, "shifted" by 1. That leads to what evail is saying:
[tex]\lim_{n\rightarrow\infty} a(n)= \sqrt{1+ \lim_{x\rightarrow\infty} a(n+1)}[/tex]
so if the limit is "[itex]l[/itex]", [itex]l= \sqrt{1+ l}[/itex] which has the solutions evail shows.

Of course, the limit is one of [itex](1+ \sqrt{5})/2[/itex] or [itex](1- \sqrt{5})/2[/itex]. Do you see how to determine which one? a(1)= 1>0 and it is easy to see that this sequence is increasing. The limit can't be negative!

Thanks HallsofIvy for completing my solution! i forgot that a(n+1)>a(n) for all n>1. hahaha.. so i had to reject the negative part :)